Multiplication of the Twisted De Rham Cohomology

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Let $\Bbbk$ be a field with characteristic $0$, $A=\Bbbk[x_1,\cdots, x_n,y_1,\cdots, y_r]$ be the Cox ring of the projective bundle $\mathbb{P}(\mathcal{E})$ over the projective toric variety $X_\Sigma$. $A$ is given with $\deg_1$ and $\deg_2$ be the degree function on $A$ satisfying $$ \deg_2 x_i = 0,\ \deg_2 y_i = 1. $$ $S$ is a homogeneous element in $S$ satisfying $\deg_1 S =0$, $\deg_2 S = 1$. We are interested in the twisted De Rham complex $(\Omega_A, d+dS)$. Unfortunately, $(\Omega_A, \wedge, d+dS)$ is not commutative differential graded algebra, or CDGA, because $$ (d+dS)(\alpha \wedge \beta) \neq (d+dS) \alpha \wedge \beta + (-1)^H{|\alpha|} \alpha \wedge (d+dS)\beta. $$

I claim that there is a multiplcation on $H(\Omega_A,d+dS)$.

Lemma 1. For each nonzero $\lambda\in \Bbbk$, we have an $A$-linear isomorphism of a cochain complex $$ \phi_{\lambda} : (\Omega_{A}, d+dS) \to (\Omega_{A}, d+\lambda dS). $$

proof. This map is defined by the automorphism $A\to A$ which maps $y_i$ to $\lambda y_i$ and maps $x_i$ to itself. This map naturally induces $\phi_{\lambda}:\Omega_{A}\to \Omega_{A}$. If $\omega\in \Omega_A^l$ is given, and with respect to $\deg_2$, it can be written as the finite summation of its homogeneous part $$ \omega = \omega_0 + \omega_1 + \cdots + \omega_d, $$ where $\deg_2 \omega_i = i$. Then $\phi_\lambda$ maps $\omega$ to $$ \phi_\lambda (\omega) = \omega_0 + \lambda \omega_1 + \cdots + \lambda^d \omega_d. $$ Since $d$ is degree preserving, $d \phi_\lambda = \phi_\lambda d$. $\wedge$ is also degree preserving, so $$ \phi_\lambda(\omega\wedge \eta) = \phi_\lambda(\omega)\wedge \phi_\lambda(\eta).$$ Especially, since $dS$ is homogeneous with $\deg_2=1$, $$ \phi_\lambda(dS\wedge \omega) = \lambda dS \wedge \phi_\lambda(\omega). $$ Hence we have $$ \phi_\lambda (d+dS) = (d+\lambda dS) \phi_\lambda. $$ This is an isomorphism since $\phi_{1/\lambda}$ is the inverse. □

Lemma 2. Suppose we have two 1-forms $\omega$ and $\eta$ in $\Omega^1_A$. Then we have a bilinear map $$ (\Omega_A, d+ \omega\wedge) \otimes (\Omega_A,d+\eta\wedge) \to (\Omega_A, d+(\omega+\eta)\wedge) $$ of cochain complices which maps $\alpha \otimes \beta$ to $\alpha \wedge \beta$.

proof. Bilinearlity is obvious since it is just the wedge product. We only need to prove that it commutes with the differential. \begin{align*} (d+(\eta+\omega)\wedge)(\alpha \wedge \beta) &= d\alpha + \omega \wedge \alpha \wedge \beta + (-1)^{|\alpha|} (\alpha \wedge d\beta + \alpha \wedge \eta \wedge \beta) \\&= (d+\omega\wedge) \alpha \wedge \beta + (-1)^{|\alpha|} \alpha \wedge (d+\eta\wedge)\beta. \end{align*} □

I claim that there is a CDGA structure on $\Omega_A^\bullet \otimes_\Bbbk \Bbbk[z]$, where the differential is given by $$ D(\omega\otimes z^k)=((d+kdS)\omega)\otimes z^k. $$

Proposition 3. $(\Omega_A^\bullet\otimes_\Bbbk \Bbbk[z],D,\wedge)$ is CDGA, where the multiplication $\wedge$ is given by $$ (\omega\otimes z^k)\wedge (\eta\otimes z^l) = (\omega\wedge\eta)\otimes z^{k+l}, $$ extended linearly.

proof. Commutativity and associativity are obvious. The differential commutes with the multiplication: \begin{align*} D((\omega\wedge\eta)\otimes z^{k+l}) &= ((d+(k+l)dS)\omega\wedge\eta )\otimes z^{k+l}\\ &=((d+kdS)\omega)\wedge \eta \otimes z^{k+l} + (-1)^{|\omega|}\omega\otimes((d+ldS)\eta)\otimes z^{k+l} \\ &=((d+kdS)\omega \otimes z^k)\wedge (\eta \otimes z^l) \\& \qquad \qquad+ (-1)^{|\omega|} (\omega \otimes z^k)\wedge ((d+ldS)\eta \otimes z^l) \\ &= D(\omega\otimes z^k)\wedge (\eta \otimes z^l) + (-1)^{|\omega|} (\omega\otimes z^k)\wedge D(\eta \otimes z^l). \end{align*} □

The cohomology of this CDGA is the direct sum of the cohomology of $(\Omega^\bullet_A,d+dS)$, \begin{align*} &H^l(\Omega_A^\bullet\otimes_k k[z],D) \\ &= \bigoplus_{k=0}^\infty H^l(\Omega_A,d+kdS)\otimes z^k \\ &= H^l(\Omega_A,d)\oplus \bigoplus_{k=1}^\infty H^l(\Omega_A,d+kdS)\otimes z^k \\ &\cong H^l(\Omega_A,d)\oplus \bigoplus_{k=1}^\infty H^l(\Omega_A,d+dS) \otimes z^k, \end{align*} where the last isomorphism is given by the direct sum of $\phi_{1/k}$ defined in Lemma 0.1.

In particular, when $l=0$, $$ H^0(\Omega_A^\bullet\otimes_k k[z],D)=\Bbbk. $$

Proposition 4. There is a (grade-)commutative multiplication $m$ on $\Bbbk\oplus H(\Omega_A,d+dS)$ which is defined as $$ m(\omega,\eta) = \phi_{\frac{|\omega|}{|\omega|+|\eta|}}(\omega)\wedge\phi_{\frac{|\eta|}{|\omega|+|\eta|}}(\eta), $$ for the homogeneous element $\omega,\eta \in H(\Omega_A,d+dS)$, and for $a\in \Bbbk$ and $\omega \in k\oplus H(\Omega_A,d+dS)$, $$ m(a,\omega) = a\omega. $$

proof. This is done by identifying $\Bbbk\oplus H(\Omega_A,d+dS)$ with the subring $$ \Bbbk\oplus \bigoplus_{k=1}^\infty H^k(\Omega_A,d+kdS)\otimes z^k $$ inside the ring $$ H(\Omega_A\otimes \Bbbk[z],D). $$ □

Remark 5. I am not sure if this is inherited from the multiplication of $\Omega_A$. As far as I know, the multuiuplication $m$ defined as $$ m(\omega,\eta) = \phi_{\frac{|\omega|}{|\omega|+|\eta|}}(\omega)\wedge\phi_{\frac{|\eta|}{|\omega|+|\eta|}}(\eta) $$ over $\Omega_A$ is commutative and associative, but does not commute with the differential, $$ (d+dS)m(\omega,\eta) \neq m((d+dS)\omega,\eta) + (-1)^{|\omega|}m(\omega,(d+dS)\eta). $$

Remark 6. The definition seems to be problematic when $\Bbbk$ is not characteristic zero.